# How to find the inverse of a function

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1. Apr 12, 2016

### Kupkake303

• moved into h/w help, so template is missing
T(t) = Ts+(98.6 – Ts)e-kt

rewrite in the form t=g-1(T)

In trying to understand how to find the inverse of this but am having a hard time, please advise.

Thanks,
Kupkake303

Last edited by a moderator: Apr 13, 2016
2. Apr 12, 2016

### stevendaryl

Staff Emeritus
This is just a matter of trying to get $t$ by itself on one side of an equality. You initially have: $T = T_s + (98.6 - T_s) e^{-kt}$. To simplify the expression, let $U = e^{-kt}$. Then the equation is: $T = T_s + (98.6 - T_s) U$. So solve for $U$. Then solve for $t$ in terms of this value of $U$.

3. Apr 13, 2016

### Kupkake303

So I would put it as 0=Ts+(98.6-Ts)U ??

Doing that would give me -Ts/(98.6-Ts)=U

Then putting the e-kt back in you get -Ts/(98.6-Ts)=e-kt

So then I would then need to get that by it's self so the t is by its self but I can't remember how to do this. Please advise. Thank you

4. Apr 13, 2016

### stevendaryl

Staff Emeritus
No, $T = T_s + (98.6 - T_s)U$

So $U = (T-T_s)/(98.6 - T_s)$

5. Apr 13, 2016

### Kupkake303

Right, because the T wouldn't just disappear.

So next I would change the U back to a e-kt for e-kt=(T-Ts)/(98.6-Ts)

from there would I try to figure out the e-kt ?

6. Apr 13, 2016

### stevendaryl

Staff Emeritus
You want to take the natural log of both sides of the equation.

7. Apr 13, 2016

### Kupkake303

okay so I would
e-kt=(T-Ts)/(98.6-Ts)

lne-kt=ln(T-Ts)/(98.6-Ts)

The ln would cancel out the e

-kt=ln(T-Ts)/(98.6-Ts)

then divide by -k? to get

t=(1/-k)ln(T-Ts)/(98.6-Ts)....?

so this would be my answer for finding the inverse ..?

8. Apr 13, 2016

### stevendaryl

Staff Emeritus
Yes, except that $-ln(A/B) = ln(B/A)$, so you can get rid of the $-$ sign to get:

$t = \frac{1}{k} ln((98.6 - T_s)/(T-T_s))$

9. Apr 13, 2016

### Kupkake303

Awesome! Thank you soooo much. I spent about 4-5 hours trying to figure this problem out myself and I wouldn't have gotten through it by myself.

10. Apr 13, 2016

### epenguin

Logarithms are the inverse of the finish of exponentiation, by definition. Hopefully you remember your first lesson about them. Well hopefully they told you that then.